There are four main problems with your function implementation.
The first one is that the function accepts the pointer to the source string by value. That is the parameter cad
is initialized by the value of the pointer used as an argument. As a result changing the variable cad
does not influence on the original pointer.
The second one is that you are not checking whether a call of realloc
was successful. As a result the function can invoke undefined behavior.
The third one is that it is inefficient to reallocate the string each time when a new character is appended.
And at last the fourth one is that the result dynamically allocated array does not contain a string because you forgot to append the terminating zero character '\0'
.
If you want to change within the function a value of the original pointer you should either to return from the function the result pointer obtained in the function and assign it to the original pointer in the caller. Or you should pass the original pointer to the function by reference. In C passing by reference means passing an object (that can be a pointer) indirectly through a pointer to it.
Here is a demonstrative program that shows the function implementation when the original pointer is accepted by the function by reference.
The function also returns a pointer to the result string that can be checked in the caller whether the reallocation of dynamic memory within the function was successful.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * remove_char( char **s, char c )
{
char * result = *s;
if ( c != '\0' )
{
char *dsn = *s;
const char *src = *s;
do
{
if ( *src != c )
{
if ( dsn != src )
{
*dsn = *src;
}
++dsn;
}
} while ( *src++ );
char *tmp = realloc( *s, ( dsn - *s ) * sizeof( char ) );
if( tmp != NULL ) *s = tmp;
result = tmp;
}
return result;
}
int main(void)
{
char *s = malloc( 12 );
strcpy( s, "H#e#l#l#o!" );
puts( s );
if ( remove_char( &s, '#' ) ) puts( s );
free( s );
return 0;
}
The program output is
H#e#l#l#o!
Hello!
Another approach is to write a function that does not change the source string but creates dynamically a new string that contains the source string excluding the specified character. Such a function is more flexible because you can call it with string literals. If the source string also was dynamically allocated then the caller of the function after a successful call it can just free the source string.
Here is a demonstrative program.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * remove_copy( const char *s, char c )
{
size_t src_len = strlen( s );
size_t dsn_len = src_len;
if ( c != '\0' )
{
for ( const char *p = s; ( p = strchr( p, c ) ) != NULL; ++p )
{
--dsn_len;
}
}
char *result = malloc( ( dsn_len + 1 ) * sizeof( char ) );
if ( result != NULL )
{
const char *src_s = s;
char *dsn_s = result;
if ( dsn_len != src_len )
{
for ( const char *p = src_s;
( p = strchr( src_s, c ) ) != NULL;
src_s = p + 1 )
{
if ( p - src_s != 0 )
{
memcpy( dsn_s, src_s, p - src_s );
dsn_s += p - src_s;
}
}
}
strcpy( dsn_s, src_s );
}
return result;
}
int main(void)
{
char s[] = "H#e#l#l#o!";
puts( s );
char *p = remove_copy( s, '#' );
if ( p != NULL ) puts( p );
free( p );
return 0;
}
The program output is the same as shown for the preceding demonstrative program that is
H#e#l#l#o!
Hello!